$\overline{AC} = 10$ $\overline{BC} = {?}$ $A$ $C$ $B$ $10$ $?$ $ \sin( \angle BAC ) = \frac{4\sqrt{41} }{41}, \cos( \angle BAC ) = \frac{5\sqrt{41} }{41}, \tan( \angle BAC ) = \dfrac{4}{5}$
Solution: $\overline{BC}$ is the opposite to $\angle BAC$ $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the adjacent side and need to solve for the opposite side so we can use the tan function (TOA) $ \tan( \angle BAC ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\overline{BC}}{\overline{AC}}= \frac{\overline{BC}}{10} $ $ \overline{BC}=10 \cdot \tan( \angle BAC ) = 10 \cdot \dfrac{4}{5} = 8$